Question

Points A and B are situated perpendicular to the axis of a 2cm long bar magnet at large distances X and 3X from its centre on opposite sides. The ratio of the magnetic fields at A and B will be approximately equal to [CPMT 1988]
A) 1 : 9 B) 2 : 9 C) 27 : 1 D) 9 : 1

Answer 1

B proportional to 1/x cube

B1/B2 = (3X) cube / (X) cube

B1:B2 = 27:1

Answer 2

The ratio of the magnetic fields at A and B will be approximately equal to

C) 27 : 1

We know, the magnetic field strength (B) of the bar magnet is given as:

B = (μ₀/2П) × (M/d³)

(μ₀/2П)×M is a constant and d is the distance from centre.

So, we represent the constant with k.

B = k/d³

⇒ B ∝ 1/d³

For A, magnetic field Ba would be

Ba = k/da³                                     [Where da is the distance of A from centre.]

Similarly, for B, magnetic field Bb would be

Bb = k/db³                                    [Where db is the distance of B from centre.]

So, Ba/Bb = db³/da³

Given db = 3X and da = 1X

Thus, Ba/Bb = (3X)³/(X)³ = 27X³/X³ = 27/1

Ratio = 27:1

Option (C) is correct.